I need to prove the following theorem:

Let $V$ be a vector space over $F$. Let $v_1,v_2,...,v_r \in V$ and $r > dim(V)$. Then, $(v_1,v_2,...,v_r)$ is linearly dependent.

My Proof Attempt:

Let dim(V) = n, for convenience, and let $(u_1,u_2,...,u_n)$ be a basis for V. Clearly, $(v_1,v_2,....,v_r)$ cannot be a basis for V since r>n and all bases have the same length.

Now, all the vectors in $(v_1,v_2,...,v_r)$ can be written as linear combinations of the basis vectors, which generate V. Hence:

$L(v_1,v_2,...,v_r) = L(u_1,u_2,....,u_n) = V$

Since $(v_1,v_2,...,v_r)$ generates V but is not a basis, it follows that it cannot be linear independent. Hence, it is linearly dependent.

The book gives a proof that is based off of the Basis Extension Theorem and I understand that proof. I was just wondering if my approach is valid or not, since I attempted to prove the result before I looked at the one given by the book.


1 Answer 1


Your equality $L(v_1,v_2,...,v_r) = L(u_1,u_2,....,u_n)$ is not true. The point is that $v_i$ is a linear combination of the $e_j's$ , so you have $\subset$, but there is no guarantee that any $e_j$ may be written as a linear combination of the $v_i's$, because there is no reason why $(v_1,\ldots,v_r)$ should be a generating family of $V$.

Think about the case $v_1=v_2=\cdots=v_r$ and $\dim_F(V)\geq 2$ for example.

  • $\begingroup$ Hmm but that seems to be a non-issue though? If $v_1 = v_2 = v_3 = .... = v_r$, then you're really just considering a list of length 1 and the dimension of your vector space is just 0, by virtue of the condition that r>n? $\endgroup$ Feb 13, 2020 at 16:01
  • $\begingroup$ No. I can use repetition, since you are talking about families of vectors. If you prefer, you could take $v_1=v, v_2=2v,\ldots,v_r=rv$ for some non zero $v$. Anyway, your argument is incorrect, for the reason i mentioned. $\endgroup$
    – GreginGre
    Feb 13, 2020 at 16:07
  • $\begingroup$ Hmm okay, thanks. $\endgroup$ Feb 13, 2020 at 16:17

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